4^17=2^34
So, 4^17-2^28=2^34-2^28=2^28(2^6-1)=2^28*63=2^28*3^2*7
You can see the biggest prime factor is 7 compared to 2 or 3
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prime factor - gmatprep
Source: Beat The GMAT — Problem Solving |
As a general strategy for exponent questions always try to reduce the terms either to lowest possible bases or to equal bases.
e.g. 4^4 - 16^2 ?
(This question I just made to illustrate the point )
you can do two things...
1) 16^2 = (4^2)^2 = 4^4
2) 4^4 = (2^2)^4 = 2^8
16^2 = (2^4)^2 = 2^8
I hope it helps.
Thanks
Amit
e.g. 4^4 - 16^2 ?
(This question I just made to illustrate the point )
you can do two things...
1) 16^2 = (4^2)^2 = 4^4
2) 4^4 = (2^2)^4 = 2^8
16^2 = (2^4)^2 = 2^8
I hope it helps.
Thanks
Amit
here it is..umaa wrote:i don't understand the answer. can you pls explain me.
4^17 -2^28
we know we can write 4^17 as (2^2)17= 2^34
thus the whole thing now is
2^34- 2^28
take 2^28 as common
2^28( 2^6-1)
which is
2^28*63 ( 2^6= 64)
now prime factors of 63 are 3^2*7
thus we know that 7 is the highest prime factor.. hope its clear.
do let me know if u have any doubts..
i will try to explain u.....dferm wrote:This is the part that is confusing me....
2^28( 2^6-1)
we know its
2^34-2^28
2^34 can also be written as 2^28*2^6 ( a^m*a*n = a^m+n )
therefore its now
2^28*2^6 - 2^28
Now we can take out 2^28 as common
2^28 ( 2^6-1) just as
15 - 5 can also be written as 5 ( 3-1)
hope it clears ur doubts ... do let me know if u still have doubts..
